# Stats Modeling The World 3rd Edition By David E. Bock – Test Bank

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1)

Suppose that after the study described in #5 we want to see if there’s evidence that the exercise

program’s effectiveness in lowering blood pressure depends on how high the person’s initial blood

pressure was. We should do a 1) A) Λ

2 goodness-of-fit test B) linear regression t-test C) matched pairs t-test D) Λ

2 test of independence E) 2-sample t-test Answer: B

Explanation: A) B) C) D) E) 2)

Several volunteers engage in a special exercise program intended to lower their blood pressure. We

measure each person’s initial blood pressure, lead them through the exercises daily for a month,

then check blood pressures again. To see if the program lowered blood pressure significantly we

should do a 2) A) 2-sample t-test B) Λ

2 test of homogeneity C) matched pairs t-test D) Λ

2 goodness-of-fit test E) linear regression t-test Answer: C

Explanation: A) B) C) D) E)

3) Create and interpret a 95% confidence interval for the slope of the regression line. 3) Answer:

Degrees of freedom = 15 – 2 = 13

95% confidence interval for Ά1 is: b1 ± t*13 × SE(b1) = 7.397 ± (2.16)(1.087) or (5.05,

9.74)

We are 95% confident that the ACT score increases an average of between 5.05 and

9.74 points for every additional GPA point. Explanation:

7)

Could eye color be a warning signal for hearing loss in patients suffering from meningitis?

British researcher Helen Cullington recorded the eye color of 130 deaf patients, and noted

whether the patient’s deafness had developed following treatment for meningitis. Her data

are summarized in the table below. Test an appropriate hypothesis and state your

conclusion.

7) Answer:

_

H0: Deafness and eye color are independent.

HA: There is an association between deafness and eye color.

These are counts of categorical data, assumed to be representative of deaf patients in

Britain, with expected counts (25.1, 76.9, 6.9, and 21.1) all at least 5. OK to do a Λ

2

test for independence, with

df = 1.

Λ

2 =

(Obs – Exp)

2

Exp =

(30 – 25.1)2

25.1 +

(72 – 76.9)2

76.9 +

(2 – 6.9)2

6.9 +

(26 – 21.1)2

21.1 = 5.87

P = 0.015

Since P = 0.015 is low, I reject the null hypothesis. There is strong evidence that

hearing loss is associated with eye color. It appears that people with dark-colored

eyes are at less risk of suffering deafness from meningitis. Explanation: 8)

Height and weight Last fall, as our first example of correlation, we looked at the heights

and weights of some AP* Statistics students. Here are the scatterplot, the residuals plot, a

histogram of the residuals, and the regression analysis for the data we collected from the

males. Use this information to analyze the association between heights and weights of

teenage boys.

9)

As part of a survey, students in a large statistics class were asked whether or not they ate

breakfast that morning. The data appears in the following table:

Is there evidence that eating breakfast is independent of the student’s sex? Test an

appropriate hypothesis. Give statistical evidence to support your conclusion. 9) 9

Answer:

_

We want to know whether the categorical variables “eating breakfast” and “student’s

sex” are statistically independent.

H0: Eating breakfast and student’s sex are independent.

HA: There is an association between eating breakfast and student’s sex.

Conditions:

*Counted data: We have the counts of individuals in categories of two categorical

variables.

*Randomization: We have a convenience sample of students, but no reason to

suspect bias.

*Expected cell frequency: The expected values (shown in parenthesis in the table)

are all greater than 5, so the condition is satisfied.

Under these conditions, the sampling distribution of the test statistic is Λ

2 with

(r – 1)(c – 1) = (2 – 1)(2 – 1) = 1 degree of freedom, and we will perform a chi-square

test of independence

11)

Voter registration A random sample of 337 college students was asked whether or not they

were registered to vote. We wonder if there is an association between a student’s sex and

whether the student is registered to vote. The data are provided in the table below

(expected counts are in parentheses). (All the conditions are satisfied – don’t worry about

checking them.)

The calculated statistic is Λ

2 = 0.249.

a. Write appropriate hypotheses.

b. Suppose the expected values had not been given. Show exactly how to calculate the

expected number of men who are registered to vote.

c. Show how to calculate the component of Λ

2 for the first cell.

d. How many degrees of freedom are there?

e. Find the P-value for this test.

f. State your complete conclusion in context. 11) Answer:

a. H0: Voter registration is independent of a student’s sex.

HA: There is an association between voter registration and a student’s sex.

b. 251

337 137 = 102

c.

(104 – 102)2

102

d. df = (2 – 1)(2

e. 0.618

f. Since the P-value of 0.618 is high, we fail to reject the null hypothesis. There is no

evidence of an association between a student’s sex and whether the student is

registered to vote.

13)

College admissions According to information from a college admissions office, 62% of the

students there attended public high schools, 26% attended private high schools, 2% were

home schooled, and the remaining students attended schools in other countries. Among

this college’s Honors Graduates last year there were 47 who came from public schools, 29

from private schools, 4 who had been home schooled, and 4 students from abroad. Is there

any evidence that one type of high school might better equip students to attain high

academic honors at this college? Test an appropriate hypothesis and state your conclusion. 13) Answer:

_

H0: Distribution of school type among honors grads is the same as for whole

college.

HA: Distribution of school type among honors grads is different.

These are counts; we assume this group is representative of other years; after

combining home schoolers and students from abroad as “other”, expected counts of

52.08, 21.84, and 10.08 are all L 5. OK to do a chi-square goodness-of-fit test with 2

df.

Λ

2 =

(Obs – Exp)

2

Exp =

(47 – 52.08)2

52.08 +

(29 – 21.84)2

21.84 +

(8 – 10.08)2

10.08 = 3.27

P = 0.195. With such a large P-value we do not reject the null hypothesis. There is no

evidence that students who graduate with honors came from different high school

backgrounds than others

17)

Peanut M&Ms According to the Mars Candy Company, peanut M&M’s are 12% brown,

15% yellow, 12% red, 23% blue, 23% orange, and 15% green. On a Saturday when you have

run out of statistics homework, you decide to test this claim. You purchase a medium bag

of peanut M&M’s and find 39 browns, 44 yellows, 36 red, 78 blue, 73 orange, and 48 greens.

Test an appropriate hypothesis and state your conclusion. 17) Answer:

_

We want to know if the distribution of colors in the bag matches the distribution

stated by the Mars Candy Company.

H0: The distribution of colors in the bag matches the distribution stated by the Mars

Candy Company.

HA: The distribution of colors in the bag does not match the distribution stated by

the Mars Candy Company.

Conditions:

*Counted data: We have the counts of the number of peanut M&Ms of each color.

*Randomization: We will assume that each bag of peanut M&Ms represents a

random

sample of peanut M&Ms.

*Expected cell frequency: There are a total of 318 peanut M&Ms. The smallest

percentage of any particular color is 12% (brown and red), and we expect 318(0.12) =

38.16. Since the smallest expected count exceeds 5, all expected counts will exceed 5,

so the condition is satisfied.

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